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3 years ago

The conjugate pair for a strong base is -

Chemistry

1 answer:

xxTIMURxx [149]3 years ago

4 0

Weak acid is the answers

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Two unknown molecular compounds were being studied. A solution containing 5.00 g of compound A in 100. g of water froze at a low

LenaWriter [7]

Answer:

Compound B has greater molar mass.

Explanation:

The depression in freezing point is given by ;

$\Delta T_f=i\times k_f\times m$ ..[1]

$m=\frac{\text{Mass of solute}}{\text{Molar mass of solute}\times \text{Mass of solvent in kg}}$

Where:

i = van't Hoff factor

$k_f$ = Molal depression constant

m = molality of the solution

According to question , solution with 5.00 g of A in 100.0 grams of water froze at at lower temperature than solution with 5.00 g of B in 100.0 grams of water.

The depression in freezing point of solution with A solute: $\Delta T_{f,A}$

Molar mass of A = $M_A$

The depression in freezing point of solution with B solute: $\Delta T_{f,B}$

Molar mass of B = $M_B$

$\Delta T_{f,A}>\Delta T_{f,B}$

As we can see in [1] , that depression in freezing point is inversely related to molar mass of the solute.

$\Delta T_f\propto \frac{1}{\text{Molar mass of solute}}$

$M_A$

This means compound B has greater molar mass than compound A,

4 0

3 years ago

Answer the question below. Use the rubric in the materials for help if needed.

earnstyle [38]

Answer:

...

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Explanation:

hop it helps.....

8 0

3 years ago

(a) The original value of the reaction quotient, Qc, for the reaction of H2(g) and I2(g) to form HI(g) (before any reactions tak

Taya2010 [7]

Answer:

Here's what I get

Explanation:

Assume the initial concentrations of H₂ and I₂ are 0.030 and 0.015 mol·L⁻¹, respectively.

We must calculate the initial concentration of HI.

1. We will need a chemical equation with concentrations, so let's gather all the information in one place.

H₂ + I₂ ⇌ 2HI

I/mol·L⁻¹: 0.30 0.15 x

2. Calculate the concentration of HI

$Q_{\text{c}} = \dfrac{\text{[HI]}^{2}} {\text{[H$_{2}$][I$_{2}$]}} =\dfrac{x^{2}}{0.30 \times 0.15} = 5.56\\\\x^{2} = 0.30 \times 0.15 \times 5.56 = 0.250\\x = \sqrt{0.250} = \textbf{0.50 mol/L}\\\text{The initial concentration of HI is $\large \boxed{\textbf{0.50 mol/L}}$}$